Problem
GEO-B2-M01-P013 Cyclicity Gives Similarity
#13
★★★★☆ Level 4 of 5
In triangle \(ABC\), points \(D\) and \(E\) lie on \(AB\) and \(AC\). It is known that \(B,C,D,E\) lie on one circle. Prove that \(\triangle ADE\sim\triangle ACB\), and derive \(AD\cdot AB=AE\cdot AC\).
First prove equality of two pairs of angles, then write the proportion of corresponding sides.
From cyclicity of \(B,C,D,E\), we get \(\angle ADE\equiv\angle ACB\) and \(\angle AED\equiv\angle ABC\). Therefore \(\triangle ADE\sim\triangle ACB\). The corresponding sides give \(\frac{AD}{AC}=\frac{AE}{AB}\). Cross-multiplying, we obtain \(AD\cdot AB=AE\cdot AC\).
This is a gentle bridge to power of a point, but the solution remains angle-based and uses similarity.